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Attention

This question has been slightly modified!!

Reference

It is related to: Bochner Integral: Axioms

Problem

Given a measure space $\Omega$ and a Banach space $E$.

Consider Bochner measurable functions $F\in\mathcal{B}$.

Then integrability is given by: $$\int\|F-S_n\|\mathrm{d}\mu\to0\iff\int\|F\|\mathrm{d}\mu<\infty$$

On the one hand it holds: $$\int\|F\|\mathrm{d}\mu\leq\int\|F-S_N\|\mathrm{d}\mu+\int\|S_N\|\mathrm{d}\mu<1+\infty$$

What about the converse?

Addendum

There's another didactic definition of integrability: $$F\in\mathcal{L}:\iff\int\|S_m-S_n\|\mathrm{d}\mu\to0\quad(S_n\to F)$$ Certainly, one has for a suitable approximation: $$S_n\to F:\quad\int\|S_m-S_n\|\mathrm{d}\mu\leq\int\|F-S_m\|\mathrm{d}\mu+\int\|F-S_n\|\mathrm{d}\mu\to0$$ But what about the converse here?

Caution

Although an integral gives the impression of measurability one should keep in mind that: $$\int\|F-S_n\|\mathrm{d}\mu\to0\quad\nRightarrow\quad F\in\mathcal{B}$$ (For a counterexample see: Bochner Integral: Approximability)

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  • $\begingroup$ First, you will need some assumptions on $f$, e.g. Bochner-measurability. (Otherwise, $\Vert f\Vert$ could be measurable without $f$ being measurable). Second, what is your definition of the Bochner-integral? $\endgroup$
    – PhoemueX
    Sep 1, 2014 at 16:22
  • $\begingroup$ @PhoemueX: Let $f$ be Bochner measurable resp. Bochner integrable if there exists a sequence resp. $L_1$-cauchy sequence of simple functions converging pointwise a.e. $s_n(\omega)\to f(\omega)$. $\endgroup$ Sep 1, 2014 at 16:56
  • $\begingroup$ @PhoemueX: Ah I get your point - will add it to the question. $\endgroup$ Sep 1, 2014 at 17:21

3 Answers 3

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You can find a clear proof in chapter IV of J. Diestel's Sequences and series in Banach spaces (p. 26).

Here is another proof.

Denote by $(\Omega, \Sigma, \mu)$ the underlying probability space, and by $X$ the involved Banach space. I guess that you implicitely assume that the function $f:\Omega\to X$ is Bochner-measurable.

Assume that $\Vert f\Vert$ is integrable. We want to show that there is a sequence of simple functions $(s_n)$ such that $\int_\Omega \Vert s_n-f\Vert\, d\mu\to 0$. Equivalently, given $\varepsilon >0$, we need to find a simple function $s$ such that $\int_\Omega \Vert s-f\Vert\leq\varepsilon$.

Since $f$ is Bochner-measurable, there is a sequence of simple functions $(\phi_n)$ converging almost everywhere to $f$. If we set $A_n=\{ x\in\Omega;\; \Vert \phi_n(x)-f(x)\Vert\geq\varepsilon\}$, then $\mu(A_n)\to 0$.

Since $\Vert f\Vert$ is integrable, one can find $\delta>0$ such that $\int_A \Vert f\Vert\, d\mu<\varepsilon$ for any measurable set $A$ satisfying $\mu(A)<\delta$. Choose $n$ such that $\mu(A_n)<\delta$, and define $s:\Omega\to X$ by $s=0$ on $A_n$ and $s=\phi_n$ outside $A_n$. This is a simple function, and $$ \int_\Omega \Vert s-f\Vert=\int_{A_n}\Vert f\Vert\, d\mu+\int_{\Omega\setminus A_n}\Vert \phi_n-f\Vert\leq \varepsilon+\varepsilon \, ,$$ where the first $\varepsilon$ comes from the fact that $\mu(A_n)<\delta$, and the second one from the fact that $\Vert \phi_n-f\Vert<\varepsilon$ on $\Omega\setminus A_n$. So we are done, upon replacing $\varepsilon$ by $\varepsilon/2$ in the beginning.

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  • $\begingroup$ I hope you don't mind that I changed the accepted answer - I guess I found a shorter one meanwhile. Still you got my (+1). $\endgroup$ Nov 28, 2014 at 22:16
  • $\begingroup$ @Freeze_S Don't worry! By the way, I didn't know it was possible to accept an answer to one of its own questions. $\endgroup$
    – Etienne
    Nov 29, 2014 at 13:27
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Proof

Suppose $\int\|F\|\mathrm{d}\mu<\infty$.

By measurability there is a simple approximation: $$S_n\in\mathcal{S}:\quad S_n\to F$$ Bound the approximation via cutoff: $$\|S'_n\|\leq2\|F\|:\quad S'_n\to F$$ so it gets an integrable dominant: $$\int\|F-S'_n\|\mathrm{d}\mu\leq\int3\|F\|\mathrm{d}\mu<\infty$$ Concluding that the integrals vanish: $$\int\|F-S'_n\|\mathrm{d}\mu\to0$$

Addendum

Given a simple cauchy approximation one has: $$S_n\to F:\quad\int\left|\|S_m\|-\|S_n\|\right|\mathrm{d}\mu\leq\int\|S_m-S_n\|\mathrm{d}\mu\to0$$ So it is absolutely integrable: $$\int\|F\|\mathrm{d}\mu=\lim_n\int\|S_n\|\mathrm{d}\mu<\infty$$ (The hard work was done in: Amann & Escher Integral vs. Lebesgue Integral)

Shortcut

By Fatou this turns into an amazing one-liner: $$\int\|F-S_n\|\mathrm{d}\mu\leq\liminf_n\int\|S_m-S_n\|\mathrm{d}\mu\to0$$

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$(\Rightarrow)$ Since $$\int \big|\|f\|-\|s_n\|\big|d\lambda \leq \int \big\|f-s_n\big\|d\lambda \rightarrow 0$$ we have $$\left|\int \|s_n\|d\lambda-\int\|s_m\|d\lambda\right|\leq \int \big|\|s_n\|-\|s_m\|\big|d\lambda \leq \int \big|\|f\|-\|s_n\|\big|d\lambda+\int \big|\|f\|-\|s_m\|\big|d\lambda \rightarrow 0$$ then, $$ \left(\int \|s_n\| d\lambda\right)_{n\in\mathbb N}\;\text{ is a Cauchy sequence in $\mathbb R$, therefore, it is convergent sequence.}$$ Suppose that $\lim\int \|s_n\|=M$. Thus, $$\int \|f\|d\lambda\leq \int\|f-s_n\|d\lambda+\int\|s_n\|d\lambda\,,\quad\forall\,n\in\mathbb N$$ taking the limit as $n\rightarrow\infty$, $$\int \|f\|d\lambda\leq 0 + M = M<\infty$$ Note that $\|f\|$ is real number, then what we had to show is that this was Lebesgue integrable.

$(\Leftarrow)$ Let us assume that $f:X\rightarrow E$ is Bochner-measurable and $f(X)$ is separable in $E$. The above assumptions imply that there is a sequence $\{s_n\}_{n\in\mathbb N}$ of simple functions such that $s_n\rightarrow f$ and $\|s_n\|\leq2\|f\|$. We have then $\|f-s_n\|\rightarrow 0$, and $$\|f-s_n\|\leq \|f\|+\|s_n\|\leq 3\|f\|$$ Define the measurable function $g_k=\|f-s_k\|$ and note that $g_k\leq 3\|f\|$. As $\|f\|$ is integrable function, we can use the dominated convergence theorem, then $$\lim_{n\rightarrow \infty}\int \|f-s_k\| d\lambda=\lim_{n\rightarrow \infty}\int g_k d\lambda = 0$$ This shows that $f$ is Bochner-integrable. $\#$

There may be some error in the details, but I think the idea is correct.

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    $\begingroup$ Your proof would be perfect if you gave some details to explain how you can ensure the "domination" $\Vert s_n\Vert\leq 2\Vert f\Vert$. $\endgroup$
    – Etienne
    Sep 2, 2014 at 18:22
  • $\begingroup$ Really, this detail is not trivial and is a consequence of the fact that $f(X)$ is separable in $E$. Assuming this fact, let $\{y_i\,{:}\,i\in\mathbb N\}$ suck that $$f(X)=\overline{\{y_i\,{:}\,i\in\mathbb N\}}$$ $\endgroup$
    – user159420
    Sep 3, 2014 at 1:05
  • $\begingroup$ Enlarging this set by an at most countable number of vectors if necessary, we may assume that $y_i\neq 0$ for all $i\in\mathbb N$. For $i,N\in\mathbb N$ define $$\hat A_i^N = \{x\in X\;{:}\; \|f(x)\|\geq 1/N\,,\;\|f(x)-y_i\|< 1/N\}$$ $\endgroup$
    – user159420
    Sep 3, 2014 at 1:05
  • $\begingroup$ As an intersection of Borel sets the $\hat A_i^N$ are Borel sets as well. Fixing $N\in\mathbb N$, we obtain a disjoint decomposition $\{A_i^N\}_{i\in\mathbb N}$ of $\bigcup_{i\in\mathbb N}\hat A_i^N$ by setting $$A_1^N := \hat A_1^N\,,\quad A_i^N:= \hat A_i^N\setminus\left[\bigcup_{j=1}^{i-1}A_j^N\right]$$ for all $i\geq 2$. Indeed, given $x\in\bigcup_{i\in\mathbb N}\hat A_i^N$ we have $$x\in A_{i_0}^N\quad \Leftrightarrow \quad\text{$i_0$ is the smallest number such that $x\in\hat A_{i_0}^N$}$$ $\endgroup$
    – user159420
    Sep 3, 2014 at 1:06
  • $\begingroup$ Since $\{y_i\,{:}\,i\in\mathbb N\}$ is dense in $f(X)$, for each $N\in\mathbb N$ we have $$\bigcup_{i\in\mathbb N}A_i^N = \bigcup_{i\in\mathbb N}\hat A_i^N = \{x\in X\;{:}\;\|f(x)\|\geq 1/N\}$$ $\endgroup$
    – user159420
    Sep 3, 2014 at 1:06

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